Question: A curve in the plane is defined parametrically by the equations $x=\tan(2t)$ and $y=\sin\left(4t-\dfrac{\pi}{2}\right)$. Find the value of $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D $0$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=\tan(2t)$ and $y=\sin\left(4t-\dfrac{\pi}{2}\right)$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\sin\left(4t-\dfrac{\pi}{2}\right)\right)}{\dfrac{d}{dt}(\tan(2t))} \\\\ &=\dfrac{4\cos\left(4t-\dfrac{\pi}{2}\right)}{2\sec^2(2t)} \\\\ &=\dfrac{2\cos\left(4t-\dfrac{\pi}{2}\right)}{\sec^2(2t)} \gray{\text{Simplify}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= {\dfrac{\pi}{2}}$ : $\begin{aligned} &\phantom{=}\dfrac{2\cos\left(4\left({\dfrac{\pi}{2}}\right)-\dfrac{\pi}{2}\right)}{\sec^2\left(2\left({\dfrac{\pi}{2}}\right)\right)} \\\\ &=\dfrac{2\cos\left(\dfrac{3\pi}{2}\right)}{\sec^2(\pi)} \\\\ &=\dfrac{2\cdot 0}{1} \\\\ &=0 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{2}$ is $0$.